Exercise 2.7.3 (Identical congruence)

Question

Prove that $x^{14} + 12x^2 \equiv 0 \pmod {13}$ has $13$ solutions and so is an identical congruence.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

\begin{proof} 
In $\Z[x]$,
$$x^{14} + 12x^2 = x(x^{13} -x) + 13 x^2.$$
Therefore, for all $a \in \Z$, by Fermat's Theorem,
$$a^{14} + 12 a^2 = a(a^{13} -a) + 13 a^2 \equiv 0 \pmod{13}.$$
So $x^{14} + 12x^2 \equiv 0 \pmod {13}$ is an identical congruence.
\end{proof}

Exercise 2.7.3 (Identical congruence)

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Exercise 2.7.3 (Identical congruence)

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